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Range of a rational polynomial (Tomato subjective – 76)

problem: Find the set of all values of \({m}\) such that \({\displaystyle {y} = {\frac{x^2-x}{1-mx}}}\) can take all real values.

solution: \({\displaystyle {y} = {\frac{x^2-x}{1-mx}}}\)
\({\Leftrightarrow}\) \({\displaystyle{y – myx = x^2 – x}}\)
\({\Leftrightarrow}\) \({x^2 + x(my – 1) – y = 0}\)
\({\Leftrightarrow}\) \({\displaystyle{x} = {\frac{1 – my {\pm} {\sqrt{(my-1)^2+4y}}}{2}}}\)
Now \({y}\) takes all real values if discriminant \({(my-1)^2 + 4y}\) is allways \({> 0}\).
So now we have to find the all values of \({m}\) such that \({(my-1)^2 + 4y}\) \({> 0}\) for all \({y}\) \({\in}\) |R.
\({\Leftrightarrow}\) \({m^2 y^2 – 2my + 1 + 4y > 0}\)
\({\Leftrightarrow}\) \({m^2 y^2 + y(4 – 2m) + 1 > 0}\) …(i)
Now this is a equation of upside open parabola. If the discriminant is \({\le{0}}\) of equation (i) then \({(my – 1)^2 + 4y}\) will always positive.
\({\Rightarrow}\) \({\displaystyle{(4 – 2m)^2 – 4m^2} {\le {0}}}\)
\({\Rightarrow}\) \({{16m + 16} {\le {0}}}\)
\({\Rightarrow}\) \({m + 1 {le{0}}}\)
\({\Rightarrow}\) \({m {\le{- 1}}}\)
Conclusion: If \({m {\le{1}}}\), then \({\displaystyle{y = {\frac{x^2 – x}{1 – mx}}}}\) can take all the values as \({x}\) varies over |R.

August 8, 2015
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